Question

The number of integral values of $x$ for which quadratic equation $y=x^2-8x+12$ is negative is

• A. 3

Answer 1

The correct option is A 3
Given: $y=x^2-8x+12$
On comparing with standard quadratic equation $y=a{x}^2+bx+c$, we get; $a=1,b=-8,c=12$ .
Now we will find the nature of roots by finding the value of $D.$
$D=b^2-4ac=8^2-4\cdot 1\cdot 12$
$\Rightarrow D=16$

$D>0$ means we have two distinct roots. And we also have $a>0$ which means parabola will be upward opening that cuts the $x-$ axis at two distinct points.

And we also know that when a parabola is opening upward with two distinct roots then it will take negative values between the two roots.
Now we will find the roots.

We have, $y=x^2-8x+12$
$\Rightarrow y=x^2-8x+12=0$
$\Rightarrow y=x^2-6x-2x+12=0$
$\Rightarrow y=x(x-6)-2(x-6)=0$
$\Rightarrow y=(x-2)(x-6)=0$
$\Rightarrow x=2\text{or}x=6$

We have $2,6$ as roots of the equation.
Now, for $x\in (2,6)y<0.$

$\therefore$ The values for $x$ for which $y<0\text{are}3,4,5.$

$\therefore$ The number of values are $3.$